Chapter 10.1, Problem 21PSC

To determine

To prove: that $AC\cong \text{B}D$ if $AB$ is diameter of the circle and $AC$ is parallel to $BD$ .

Answer to Problem 21PSC

  $AC=\text{ B}D.$

Explanation of Solution

Given:

  

  $AC\parallel BD$ .

Concept used:

Condition of congruence triangle:

Side angle side $\left(SAS\right)$ : two side and one angle of one triangle is equivalent another triangle.

Side-side-side $\left(SSS\right)$: all three side of the one triangle is equivalent to another triangle.

Angle side angle

  $\left(ASA\right)$ : two angle and side included between the angles of one triangle is equivalent to another two angle and side included between the angles of triangle.

Angle-angle side $\left(AAS\right)$:

two angle and non-included side between the angles of one triangle is equivalent to another two angle and non-included side between the angles of triangle.

Right angle hypotenuse side $\left(RHS\right)$: Hypotenuse and side of a right-angled triangle is equivalent to the hypotenuse and side of the second right angled triangle.

Angle opposite congruent sides in the triangle are congruent.

Corresponding parts of congruent triangles are congruent $\left(CPCTC\right)$:

When triangles are congruent, one triangle can be moved to coincide with the other triangle. All corresponding sides and angles will be congruent.

Calculation:

According to the given:

  $AC\parallel BD$ .

Which implies angle $\angle 1=\angle 2$ .

Here, $AB$ is diameter of the circle.

Which implies $OA=OB=\text{radius}$ .

Join $OC\text{ and }OD$ .

Now the circle will have:

  $\begin{array}{l}OA=OC=OD=OB\mathrm{.....}\left(i\right)\\ AO=OC.\\ \Rightarrow \angle 1=\angle 2.\\ OD=OB.\\ \Rightarrow \angle 2=\angle 4.\end{array}$

Since, angle opposite congruent sides in the triangle are congruent.

  $\begin{array}{l}\angle 1=\angle 2=\angle 3=\angle 4.\\ \angle AOC={180}^0-\angle 1-\angle =3.\\ \Rightarrow {180}^0-\angle 4-\angle 2=\angle DOB\mathrm{....}\left(ii\right)\end{array}$

From $\left(i\right)\text{ and }\left(ii\right)$ :

By $SAS$ congruence criterion:

  $\Delta AOC\text{ congruent }\Delta BOD$ .

congruent parts of congruent triangles are congruent.

Hence, $AC=\text{ B}D.$